Question

The percentage change in the flat surface before and after-when a solid metal cylinder of height 10 units and radius 7 units is melted and recast into two cones in the volume proportion of $3: 4$, keeping the height same would be_____

Answer 1

Let the radius of cylinder $=R$
And that of cone 1 and cone $2=r_1$ and $r_2$
Let volume be $V, V_1$ and $V_2$ and $V_2$, respectively.
Cones are recasted from molten cylinder
So $V=V_1+V_2$
Ratio of volume of the two cones $V_1: V_2:$ :
$3: 4$
$\Rightarrow V_1=3 x$ and $V_2=4 x$
Thus, $V=V_1+V_2=3 x+4 x=7 x$
Volume of cylinder $=\pi R^2 h$
Volume of cone $1=\frac{1}{3} \pi r_1^2 h$
Volume of cone $2=\frac{1}{3} \pi r_2^2 h$
We know ratio of volume is $7: 3: 4$
$\Rightarrow \pi R^2 h: \frac{1}{3} \pi r_1^2 h: \frac{1}{3} \pi r_2^2 h $
$ R^2: \frac{1}{3} r_1^2: \frac{1}{3} r_2^2:: 7: 3: 4 $
$ R^2: r_1^2: r_2^2=7: 9: 12=k$
$R^2=7 k$
$r_1^2=9 k$
$r_2^2=12 k$
Flat surface area of cylinder is sum of the area of two circles at the top and bottom.
So, flat surface area of cylinder $=2 \pi R^2$
Flat surface area of cone 1 , and cone $2=\pi r_1^2$ and $\pi r_2^2$
Ratio of flat surface area of cylinder so that of two cone 1 and cone 2
$2 \pi R^2: \pi r_1^2+\pi r_2^2 $
$2 R^2: r_1^2+r_2^2 $
$2(7 k): 9 k+12 k$
$14 k: 21 k$
Change in flat surface area $=21 k-14 k=7 k$ $\%$ change $=\frac{7 k}{14 k} \times 100=50 \%$